2024 AIME I Problems/Problem 2

Revision as of 19:47, 2 February 2024 by Goezonme (talk | contribs) (Solution 1)

Problem

There exist real numbers $x$ and $y$, both greater than 1, such that $\log_x\left(y^x\right)=\log_y\left(x^{4y}\right)=10$. Find $xy$.

Solution 1

By properties of logarithms, we can simplify the given equation to $x\log_xy=4y\log_yx=10$. Let us break this into two separate equations: xlogxy=144ylogyx=14. We multiply the two equations to get: \[4xy\left(\log_xy\log_yx\right)=196.\]

Also by properties of logarithms, we know that $\log_ab\cdot\log_ba=1$; thus, $\log_xy\cdot\log_yx=1$. Therefore, our equation simplifies to:

\[4xy=196\implies xy=\boxed{049}.\]

~Technodoggo

Solution 2

Convert the two equations into exponents:

\[x^{10}=y^x~(1)\] \[y^{10}=x^{4y}~(2).\]

Take $(1)$ to the power of $\frac{1}{x}$:

\[x^{\frac{10}{x}}=y.\]

Plug this into $(2)$:

\[x^{(\frac{10}{x})(10)}=x^{4x^{\frac{10}{x}}}\] \[{\frac{100}{x}}={4(x^{\frac{10}{x}})}\] \[{\frac{25}{x}}={x^{\frac{10}{x}}}=y,\]

So $xy=\boxed{025}$

~alexanderruan

See also

2024 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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